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I have a very simple question about polar coordinate system. According to Wikipedia, the angle is defined as:

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where atan2(y,x) is defined as "a common variation on the arctangent function"

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Since I'm not an advanced student in calculus, I was trying to determine these values as an exercise, just by appling the inverse relation between tangent and arctangent function, and by noticing that the ratio of the argument of the arctangent function is in fact the ratio between the sinus and the cosinus of the angle. I've found correctly the values for the angle in the last three cases of the system above, but I don't understand why I should add and subtract \pi in the second and third case respectively. Isn't the period of the tangent function equal to \pi itself? Why am I wrong? I suppose that there is some advanced additional stuff that I can't understand! As usual, thank you!

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The problem is that $\frac{a}{b}$ and $\frac{-a}{b}$ result in the same value when you plug them into $tan(x)$, which means that if we are going the other way around (from $tan(x)$ to $x$), then there is no clear solution. Which of the two should be returned. This comes down to the function $tan(x)$ not being one-to-one.

Now in the $(x,y)$-plane when you know the sign of $x$ and of $y$ then you know in which quadrant they lie and thus (coming polar coordinates) in which "quarter" of the circle they are in. For example if $x$ is positive and $y$ is negative, then I know that the angle this point will have in polar coordinates will be between 270 and 360 degrees or $\frac{3\pi}{2}$ and $\frac{4\pi}{2} = 2 \pi$.

This allows me to choose the appropriate value when using $arctan(x)$ up there. So these cases are needed because $tan(x)$ is not one-to-one and they are defined using this additional knowledge about the position of a point given the signs of its components.

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  • $\begingroup$ Thank you for replying, it's clear to me that we can determine in which quarter an angle lies, but why do we need to add \pi and -\pi? Isn't it the same angle? $\endgroup$
    – Nenne
    Aug 10 '17 at 9:40
  • $\begingroup$ Ah, you mean that $angle + \pi$ and $angle - \pi$ would result in the same angle? That is right. This measure is only done so the resulting angle ends up being within $[0, 2 \pi]$. If your angle was $\frac{3\pi}{2}$ and you added $\pi$, then you would end up with $\frac{5\pi}{2}$ which would of course be equivalent to $\frac{1\pi}{2}$ for the angle, but out of the "bounds" set for the angle to be in. $\endgroup$
    – tehfurbolg
    Aug 10 '17 at 10:18
  • $\begingroup$ I was thinking that angle and angle +π should be the same thing, if the period of the tangent function is π. How does this relate with the period of the tangent? $\endgroup$
    – Nenne
    Aug 10 '17 at 10:21
  • $\begingroup$ Ah, no! $angle$ and $angle + \pi$ are different. The full circle is 360 degress or $2 \pi$, thus only $\pi$ is half a circle or 180 degrees. $\endgroup$
    – tehfurbolg
    Aug 10 '17 at 10:22
  • $\begingroup$ Ok, I got my last mistake (sorry). In fact the angles are different, but the values of the tangent of them repeat themselves after a period of π ! I don't wanna bother you, but I'm still struggling to get the first two expressions in the system above... $\endgroup$
    – Nenne
    Aug 10 '17 at 10:26
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The point is we can't use $\arctan\frac{y}{x}$ in general because $(a,\,b)$ is antipodal to $(-a,\,-b)$, which achieves the same ratio.

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  • $\begingroup$ Could you explain me your statement more in depth, please? $\endgroup$
    – Nenne
    Aug 10 '17 at 9:25
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The point of a coordinate system is that given a radius and an angle you define one and only one point in the plan.

Adding and substracting $\pi$ when $(x,y)$ lies "in the left quarters" ensures that the angle is a continuuous, increasing function to $(-\pi$ ; $\pi]$ when you rotate counter-clockwise from $(-1;0)$ along the unit circle.

Hence from a given $(r,\theta) \in (0,\infty)\times(-\pi,\pi]$ you define a single point of the plan minus its origin.

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